Consider a matrix $n\times n$ $A$ which depends on a $n\times n$ matrix $B$ i.e., $A=A(B)$. The matrix $B$ in turn depends upon a continuous real parameter $\phi$ such that for small $\phi$, $B$ can be written as $$B(\delta\phi)=\mathbb{1}-C\delta\phi$$ where $C$ is also a $n\times n$ matrix.

Can we expand $A(B(\delta\phi))=A(\mathbb{1}-C\delta\phi)$ in a Taylor series?


1 Answer 1


I would simplify your question in this way: how a matrix with entries $a_{ij}=a_{ij}(t)$ depending on a real parameter $t$ can be expanded into a Taylor series ?

There is a simple answer : expand separately all the $a_{ij}(t)$s into Taylor series (around the origin or around another point, if they can be expanded...), then factorize $1,t,t^2...$.

The best is to see it on an example :

$$\begin{pmatrix}\cos(t)&\sin(t)\\0&1\end{pmatrix}=I_2+t\begin{pmatrix}0&1\\0&0\end{pmatrix}+t^2/2\begin{pmatrix}-1&0\\ \ \ 0&0\end{pmatrix}+\cdots$$

  • $\begingroup$ Actually, I need to know the expansion of something like $A(I+C\epsilon)$ where A, C are square matrices, and I is the identity matrix, and $\epsilon$ is small. It's actually related to a physics question I asked here. physics.stackexchange.com/questions/398463/… @JeanMarie $\endgroup$
    – SRS
    Apr 8, 2018 at 8:41
  • 1
    $\begingroup$ I have had a look at your question on physics SE. I think that finding higher order terms deals with Baker-Campbell-Hausdorff formula involving brackets, brackets of brackets, etc.(en.wikipedia.org/wiki/…). Besides : what you call "Wigner's theorem" is not known in mathematics under this terminology ; it is a fact that any rotation matrix can be expressed as the exponential of a skew symmetric matrix of the same size with coefficients that are linked to a normalized axis and the rotation angle. $\endgroup$
    – Jean Marie
    Apr 8, 2018 at 8:59
  • $\begingroup$ @JeanMarie how would one generalise this expansion to for example two variables ? Say $a_{ij} = a_{ij}(t, u)$ $\endgroup$
    – papabiceps
    Jan 22, 2019 at 13:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.